A function is a continuous function on an interval if it is continuous at each and every point on that interval.  You are able to check and see if a function is continuous at certain points using the definition of continuity:

A function \(\displaystyle f(x)\) is continuous at \(\displaystyle x = a\) if \(\displaystyle lim_{x\rightarrow a} f(x) = f(a)\)

This definition assumes that both \(\displaystyle lim_{x\rightarrow a}\) and \(\displaystyle f(a)\) exist.  If either of these do not exist or the if \(\displaystyle lim_{x\rightarrow a} f(x) \neq f(a)\), then the function \(\displaystyle f(x)\) is not continuous at point  and the function is not a continuous function on any interval containing \(\displaystyle a\).

From the graph we see that at \(\displaystyle f(-1)\) there is a hole in the graph.  This means that the function is not continuous there, this type of discontinuity is called removable discontinuity.

It is important to note that rational functions are continuous everywhere except for the points where their denominator is equal to zero.  So to find the points of discontinuity we set the denominator equal to zero and solve for \(\displaystyle x\).

\(\displaystyle x^2 - 2 +1 = 0\)

\(\displaystyle (x - 1)(x - 1) = 0\)                            (factoring)

\(\displaystyle (x-1)^2 = 0\)

Since this factors into \(\displaystyle (x-1)^2\) we only need to set \(\displaystyle (x-1)\) equal to zero to solve.

\(\displaystyle x-1 = 0\)

\(\displaystyle x=1\)

So this function is continuous everywhere except for \(\displaystyle x=1\).  We are trying to determine whether this function is continuous on the given interval \(\displaystyle [-2, 2]\) since  is in this interval, then the function is not continuous in the given interval.

We must use our definition of continuity:

\(\displaystyle f(x)\) is continuous at \(\displaystyle x = 1\) if \(\displaystyle lim_{x\rightarrow 1}f(x) = f(1)\)

Now we will use this definition and work through the limit as \(\displaystyle x\) approaches \(\displaystyle 1\).

\(\displaystyle lim_{x\rightarrow 1} f(x) = lim_{x\rightarrow 1} \frac{2x+5}{x-2} = \frac{lim_{x\rightarrow 1} 2x+5}{lim_{x\rightarrow 1}x - 2} = \frac{2(1) +5}{1-2} = \frac{7}{-1} = -7\)

Now let’s check \(\displaystyle f(-1)\)

\(\displaystyle f(-1) = \frac{2(1) + 5}{1-2} = \frac{7}{-1} = -7\)

So \(\displaystyle lim_{x\rightarrow 1}f(x) = f(1)\) and therefore this function is continuous at  \(\displaystyle 1\).

It helps if we plot this function to see what is happening.

So we see that when \(\displaystyle x< 0\) the graph of the function stays at \(\displaystyle 1\), but when \(\displaystyle x\geq 0\) then the function of the graph is equal to \(\displaystyle x\).  Even though there is a solution for \(\displaystyle x = 0\), the limit does not exist.  This is called jump discontinuity, where the graph of a function (also the function itself but it is easier to consider the graph) jumps from one solution to another with a break in the graph.

We must use our definition of continuity: \(\displaystyle f(x)\) is continuous at \(\displaystyle x = 0\) if \(\displaystyle lim_{x\rightarrow 0} f(x) = f(0)\)

Now we solve for the limit as \(\displaystyle x\) approaches \(\displaystyle 0\) and we will solve for \(\displaystyle f(0)\).

\(\displaystyle lim_{x\rightarrow 0} f(x) = lim_{x\rightarrow 0}\frac{x^3-5}{x^2+4} = \frac{lim_{x\rightarrow 0}x^3-5}{lim_{x\rightarrow 0}x^2+4} = \frac{0 - 5}{0 + 4} = -\frac{5}{4}\)

Now we will solve for \(\displaystyle f(0)\)

\(\displaystyle f(0) = \frac{0 - 5}{0+4} = -\frac{5}{4}\)

And so \(\displaystyle lim_{x\rightarrow 0} f(x) = f(0)\) meaning that this function is continuous at \(\displaystyle x = 0\).

We know that from the definition of continuity \(\displaystyle lim_{x\rightarrow a} f(x) = f(a)\).  \(\displaystyle f(x)\) must be defined (i.e. exist) at point \(\displaystyle a\) in order for a limit to be equal to \(\displaystyle f(a)\) and the limit must also exist in order to be equal to \(\displaystyle f(a)\).

A function is continuous from the right if \(\displaystyle lim_{x\rightarrow a^+} f(x) = f(a)\) and a function is continuous from the left if \(\displaystyle lim_{x\rightarrow a^-} f(x) = f(a)\).  Take the graphed function below for example:

We would say that this function is right continuous at \(\displaystyle x = 0\) because this is where the jump discontinuity occurs and the point \(\displaystyle 0\) is included when approached from the right but not from the left.

This is true.  Take the binomial \(\displaystyle x^2 + 1x^2 + 1\) for example.  This function’s domain is defined on the entire real number line so it is defined at each point of the real numbers as well.  A limit exists for this function at any point and there are no breaks or jumps within this function.  So all polynomials are continuous at all real numbers.

We know that all rational functions are continuous everywhere except for when the denominator is equal to \(\displaystyle 0\).  So we will begin by setting the denominator equal to \(\displaystyle 0\) and solving for \(\displaystyle x\).

\(\displaystyle 3cos(x) - 2 = 0\)

\(\displaystyle cos(x) = \frac{2}{3}\)

\(\displaystyle x=cos^{-1}(\frac{2}{3})\)

\(\displaystyle x=0.841\)

But this only gives us a solution between \(\displaystyle 0\) and \(\displaystyle \pi\).  So we must also consider \(\displaystyle 2\pi - 0.841 = 5.442\).  We know that each of these solutions will also have discontinuities at them for each rotation of a circle which is \(\displaystyle 2\pi\).  So the function is discontinuous at:

\(\displaystyle x=0.841 + 2\pi n\) where \(\displaystyle n=\pm 1, \pm 2,\)

\(\displaystyle x=5.442 + 2\pi n\) where \(\displaystyle n=\pm 1, \pm 2,\)